PAT A 2020年冬 7-3 File Path AC代码_冬级代码-优快云博客

本文链接：https://blog.youkuaiyun.com/maxleo928/article/details/120196036

该博客主要介绍了一个在Windows文件资源管理器中查找选定文件路径的算法。输入为目录树结构和查询ID，输出为从根目录到文件的路径。通过计数缩进的空格数量来确定层级，并使用栈模拟路径回溯。给出的AC代码实现了这一功能，能正确处理不存在于树中的ID情况。

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7-3 File Path (25 分)

The figure shows the tree view of directories in Windows File Explorer. When a file is selected, there is a file path shown in the above navigation bar. Now given a tree view of directories, your job is to print the file path for any selected file.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤103), which is the total number of directories and files. Then N lines follow, each gives the unique 4-digit ID of a file or a directory, starting from the unique root ID 0000. The format is that the files of depth d will have their IDs indented by d spaces. It is guaranteed that there is no conflict in this tree structure.

Then a positive integer K (≤100) is given, followed by K queries of IDs.

Output Specification:

For each queried ID, print in a line the corresponding path from the root to the file in the format: 0000->ID1->ID2->...->ID. If the ID is not in the tree, print Error: ID is not found. instead.

Sample Input:

14
0000
 1234
  2234
   3234
    4234
    4235
    2333
   5234
   6234
    7234
     9999
  0001
   8234
 0002
4 9999 8234 0002 6666

结尾无空行

Sample Output:

0000->1234->2234->6234->7234->9999
0000->1234->0001->8234
0000->0002
Error: 6666 is not found.

结尾无空行

思路：使用cnt来记录层级（string的长度），使用vector ans来模拟堆栈，记录访问，层级不对就pop直到cnt符合

AC代码：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct node{
	int num;
	vector<int> next;
};
int N,K;
vector<int> ans;
int main(){
	cin>>N;
	getchar();
	vector<string> v(N);
	for(int i=0;i<N;i++){
		getline(cin,v[i]);
	}
	cin>>K;
	for(int i=0;i<K;i++){
		int tag;
		scanf("%d",&tag);
		int cnt=4;
		ans.clear();
		bool flag=false;
		for(int j=0;j<N;j++){
			if(stoi(v[j])==tag){
				for(int m=0;m<(cnt-v[j].size());m++){
					ans.pop_back();
				}
				ans.push_back(tag);
				flag=true;
				break;
			}else{
				if(cnt==v[j].size()){
					ans.push_back(stoi(v[j]));
					cnt++;
				}else{
					for(int m=0;m<(cnt-v[j].size());m++){
						ans.pop_back();
					}
					ans.push_back(stoi(v[j]));
					cnt=v[j].size()+1;
				}
			}
		}
		if(flag){
			for(int j=0;j<ans.size();j++){
				if(j==0){
					printf("%04d",ans[j]);
				}else{
					printf("->%04d",ans[j]);
				}
			}
			printf("\n");
		}else{
			printf("Error: %d is not found.\n",tag);
		}
	}
	return 0;
}